Prove that each term of the following sequence is a perfect square:
49 , 4489 , 444889 , 44448889............................
Your very first number is NOT a perfect square! You obviously mean 49, not 44. And then each succeeding number is gained by putting a new "4" and "8" into the center. Have you looked at the square root of each of those numbers? That should give you a hint.
Hello, geniusgarvil!
Prove that each term of the following sequence is a perfect square:
. . $\displaystyle 49,\,4489,\,444889,\,44448889\,\hdots$
$\displaystyle \text{The }4^{th}\text{ term is: }\:a_4 \;=\;4444(10^4) + 888(10) + 9$
. . . . . . . . . . . . $\displaystyle =\;4(1111)(10^4) + 8(111)(10) + 9$
. . . . . . . . $\displaystyle =\;4\left(\frac{10^4-1}{9}\right)(10^4) + 8\left(\frac{10^3-1}{9}\right)(10) + 9 $
$\displaystyle \text{The }n^{th}\text{ term is: }\:a_n \;=\;4\left(\frac{10^n-1}{9}\right)10^n + 8\left(\frac{10^{n-1}-1}{9}\right)10 + 9$
$\displaystyle \text{We have: }\:\tfrac{4}{9}(10^{2n}) - \tfrac{4}{9}(10^n) + \tfrac{8}{9}(10^n) - \tfrac{8}{9}(10) + 9 $
. . . . . $\displaystyle =\;\tfrac{4}{9}(10^{2n}) + \tfrac{4}{9}(10^n) + \tfrac{1}{9}$
. . . . . $\displaystyle =\;\tfrac{1}{9}\big[4(10^{2n}) + 4(10^n) + 1\big]$
. . . . . $\displaystyle =\;\tfrac{1}{9}\big[2(10^n) + 1\big]^2$
. . . . . $\displaystyle =\;\left[\frac{2(10^n) + 1}{3}\right]^2 \;\;\hdots\;\;\text{ a square!}$